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Life tables
Age-specific probability statistics
Force of mortality qx
Survivorship lx
ly / lx = probability of living from age x to age y
Fecundity mx
Realized fecundity at age x = lxmx
Net reproductive rate R0 = lxmx
Generation time T = xlxmx
Reproductive value vx = (lt / lx ) mt
Ex = Expectation of further life
T, Generation time = average time from one gener-
ation to the next (average time from egg to egg)
vx = Reproductive Value = Age-specific expectation of all future offspring
p.143, right hand equation“dx” should be “dt”
vx = mx + (lt / lx ) mt
Residual reproductive value = age-specific
expectation of offspring in distant future
vx* = ( lx+1 / lx ) vx+1
Illustration of Calculation of Ex, T, R0, and vx in a Stable Population with Discrete Age Classes_____________________________________________________________________
Age Expectation Reproductive Weighted of Life Value
Survivor- Realized by Realized Ex vx
Age (x) ship Fecundity Fecundity Fecundity lx mx lxmx x lxmx
_____________________________________________________________________0 1.0 0.0 0.00 0.00 3.40 1.001 0.8 0.2 0.16 0.16 3.00 1.252 0.6 0.3 0.18 0.36 2.67 1.403 0.4 1.0 0.40 1.20 2.50 1.654 0.4 0.6 0.24 0.96 1.50 0.655 0.2 0.1 0.02 0.10 1.00 0.106 0.0 0.0 0.00 0.00 0.00 0.00Sums 2.2 (GRR) 1.00 (R0) 2.78 (T) _____________________________________________________________________E0 = (l0 + l1 + l2 + l3 + l4 + l5)/l0 = (1.0 + 0.8 + 0.6 + 0.4 + 0.4 + 0.2) / 1.0 = 3.4 / 1.0E1 = (l1 + l2 + l3 + l4 + l5)/l1 = (0.8 + 0.6 + 0.4 + 0.4 + 0.2) / 0.8 = 2.4 / 0.8 = 3.0E2 = (l2 + l3 + l4 + l5)/l2 = (0.6 + 0.4 + 0.4 + 0.2) / 0.6 = 1.6 / 0.6 = 2.67E3 = (l3 + l4 + l5)/l3 = (0.4 + 0.4 + 0.2) /0.4 = 1.0 / 0.4 = 2.5E4 = (l4 + l5)/l4 = (0.4 + 0.2) /0.4 = 0.6 / 0.4 = 1.5E5 = (l5) /l5 = 0.2 /0.2 = 1.0v1 = (l1/l1)m1+(l2/l1)m2+(l3/l1)m3+(l4/l1)m4+(l5/l1)m5 = 0.2+0.225+0.50+0.3+0.025 = 1.25 v2 = (l2/l2)m2 + (l3/l2)m3 + (l4/l2)m4 + (l5/l2)m5 = 0.30+0.67+0.40+ 0.03 = 1.40 v3 = (l3/l3)m3 + (l4/l3)m4 + (l5/l3)m5 = 1.0 + 0.6 + 0.05 = 1.65 v4 = (l4/l4)m4 + (l5/l4)m5 = 0.60 + 0.05 = 0.65v5 = (l5/l5)m5 = 0.1___________________________________________________________________________
Illustration of Calculation of Ex, T, R0, and vx in a Stable Population with Discrete Age Classes_____________________________________________________________________
Age Expectation Reproductive Weighted of Life Value
Survivor- Realized by Realized Ex vx
Age (x) ship Fecundity Fecundity Fecundity lx mx lxmx x lxmx
_____________________________________________________________________0 1.0 0.0 0.00 0.00 3.40 1.001 0.8 0.2 0.16 0.16 3.00 1.252 0.6 0.3 0.18 0.36 2.67 1.403 0.4 1.0 0.40 1.20 2.50 1.654 0.4 0.6 0.24 0.96 1.50 0.655 0.2 0.1 0.02 0.10 1.00 0.106 0.0 0.0 0.00 0.00 0.00 0.00Sums 2.2 (GRR) 1.00 (R0) 2.78 (T) _____________________________________________________________________E0 = (l0 + l1 + l2 + l3 + l4 + l5)/l0 = (1.0 + 0.8 + 0.6 + 0.4 + 0.4 + 0.2) / 1.0 = 3.4 / 1.0E1 = (l1 + l2 + l3 + l4 + l5)/l1 = (0.8 + 0.6 + 0.4 + 0.4 + 0.2) / 0.8 = 2.4 / 0.8 = 3.0E2 = (l2 + l3 + l4 + l5)/l2 = (0.6 + 0.4 + 0.4 + 0.2) / 0.6 = 1.6 / 0.6 = 2.67E3 = (l3 + l4 + l5)/l3 = (error: extra terms) 0.4 + 0.4 + 0.2) /0.4 = 1.0 / 0.4 = 2.5E4 = (l4 + l5)/l4 = (error: extra terms) 0.4 + 0.2) /0.4 = 0.6 / 0.4 = 1.5E5 = (l5) /l5 = 0.2 /0.2 = 1.0v1 = (l1/l1)m1+(l2/l1)m2+(l3/l1)m3+(l4/l1)m4+(l5/l1)m5 = 0.2+0.225+0.50+0.3+0.025 = 1.25 v2 = (l2/l2)m2 + (l3/l2)m3 + (l4/l2)m4 + (l5/l2)m5 = 0.30+0.67+0.40+ 0.03 = 1.40 v3 = (l3/l3)m3 + (l4/l3)m4 + (l5/l3)m5 = 1.0 + 0.6 + 0.05 = 1.65 v4 = (l4/l4)m4 + (l5/l4)m5 = 0.60 + 0.05 = 0.65v5 = (l5/l5)m5 = 0.1___________________________________________________________________________
Stable age distribution
Stationary age distribution
Intrinsic rate of natural increase (per capita)
r = b – d
when birth rate exceeds death rate (b > d), r is positive
when death rate exceeds birth rate (d > b), r is negative
Euler’s implicit equation:
e-rx lxmx = 1
(solved by iteration)
If the Net Reproductive Rate R0 is near one,
r ≈ loge R0 /T
J - shaped exponential population growth
http://www.zo.utexas.edu/courses/THOC/exponential.growth.html
When R0 equals one, r is zero
When R0 is greater than one, r is positive
When R0 is less than one, r is negative
Maximal rate of natural increase, rmax
Instantaneous rate of change of N at time t
is total births (bN) minus total deaths (dN)
dN/dt = bN – dN = (b – d )N = rN
Nt = N0 ert (integrated version of dN/dt = rN)
log Nt = log N0 + log ert = log N0 + rt
log R0 = log 1 + rt (make t = T)
r = log or = er (is the finite rate of
increase)
Estimated Maximal Instantaneous Rates of Increase (rmax, per capita per day) and Mean Generation Times ( in days) for a Variety of Organisms___________________________________________________________________Taxon Species rmax Generation Time (T)-----------------------------------------------------------------------------------------------------Bacterium Escherichia coli ca. 60.0 0.014Protozoa Paramecium aurelia 1.24 0.33–0.50Protozoa Paramecium caudatum 0.94 0.10–0.50Insect Tribolium confusum 0.120 ca. 80Insect Calandra oryzae 0.110(.08–.11) 58Insect Rhizopertha dominica 0.085(.07–.10) ca. 100Insect Ptinus tectus 0.057 102Insect Gibbum psylloides 0.034 129Insect Trigonogenius globulosus 0.032 119Insect Stethomezium squamosum 0.025 147Insect Mezium affine 0.022 183Insect Ptinus fur 0.014 179Insect Eurostus hilleri 0.010 110Insect Ptinus sexpunctatus 0.006 215Insect Niptus hololeucus 0.006 154Mammal Rattus norwegicus 0.015 150Mammal Microtus aggrestis 0.013 171Mammal Canis domesticus 0.009 ca. 1000Insect Magicicada septendecim 0.001 6050Mammal Homo “sapiens” (the sap) 0.0003 ca. 7000
__________________________________________________________________ _
Inverserelationshipbetween rmax
and generation time, T
Demographic and Environmental Stochasticity
random walks, especially important in small populations
Evolution of Reproductive Tactics
Semelparous versus Interoparous
Big Bang versus Repeated Reproduction
Reproductive Effort (parental investment)
Age of First Reproduction, alpha,
Age of Last Reproduction, omega,
Mola mola (“Ocean Sunfish”) 200 million eggs!
Poppy (Papaver rhoeas)produces only 4 seeds whenstressed, but as manyas 330,000 under idealconditions
Indeterminant Layers
How much should an organism invest in any given act of reproduction? R. A. Fisher (1930) anticipated this question long ago:
‘It would be instructive to know not only by what physiological mechanism a just apportionment is made between the nutriment devoted to the gonads and that devoted to the rest of the parental organism, but also what circumstances in the life history and environment would render profitable the diversion of a greater or lesser share of available resources towards reproduction.’ [Italics added for emphasis.]
Reproductive Effort
Ronald A. Fisher
Asplanchna (Rotifer)
Trade-offs between present progenyand expectation of future offspring
Iteroparous organism
Semelparous organism
Patterns in Avian Clutch SizesAltrical versus Precocial
Patterns in Avian Clutch SizesAltrical versus Precocial
Nidicolous vs. NidifugousDeterminant vs. Indeterminant Layers
N = 5290 Species
Patterns in Avian Clutch SizesOpen Ground Nesters Open Bush Nesters Open Tree Nesters
Hole Nesters
MALE (From: Martin and Ghalambor 1999)
Patterns in Avian Clutch SizesClassic Experiment: Flickers usually lay 7-8 eggs, but in an egg removal experiment, a female flicker
laid 61 eggs in 63 days
Great Tit Parus major
David Lack
Parus major
European Starling, Sturnus vulgaris
Chimney Swift, Apus apus
Seabirds (N. Philip Ashmole)
Boobies, Gannets, Gulls, Petrels, Skuas, Terns, Albatrosses
Delayed sexual maturity, Small clutch size, Parental care
Albatross Egg Addition Experiment
Diomedea immutabilis
An extra chick added to eachof 18 nests a few days afterhatching. These nests with twochicks were compared to 18 othernatural “control” nests with onlyone chick. Three months later, only 5 of the 36 experimental chicks survived from the nests with 2 chicks, whereas 12 of the 18 chicks from single chick nests were still alive. Parents could not find food enough to feed two chicks and most starved to death.
Latitudinal Gradients in Avian Clutch Size
Latitudinal Gradients in Avian Clutch Size
Daylength Hypothesis
Prey Diversity Hypothesis
Spring Bloom or Competition Hypothesis
Latitudinal Gradients in Avian Clutch Size
Nest Predation Hypothesis Alexander Skutch ––––––>
Latitudinal Gradients in Avian Clutch Size
Hazards of Migration Hypothesis
Falco eleonora
Evolution of Death Rates
Senescence, old age, genetic dustbin
Medawar’s Test Tube Model
p(surviving one month) = 0.9
p(surviving two months) = 0.92
p(surviving x months) = 0.9x
recession of time of expression of the overt effects of a
detrimental allele
precession of time of expression of the effects of a
beneficial allele
Peter Medawar
Age Distribution ofMedawar’s test tubes
Percentages of people with lactose intolerance
Joint Evolution of Rates of Reproduction and Mortality
Donald Tinkle
Sceloporus
J - shaped exponential population growth
http://www.zo.utexas.edu/courses/THOC/exponential.growth.html
Instantaneous rate of change of N at time t
is total births (bN) minus total deaths (dN)
dN/dt = bN – dN = (b – d )N = rN
Nt = N0 ert (integrated version of dN/dt = rN)
log Nt = log N0 + log ert = log N0 + rt
log R0 = log 1 + rt (make t = T)
r = log or = er (is the finite rate of
increase)
Once, we were surrounded by wilderness and wild animals,But now, we surround them.
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